# Thread: Trignometry - modelling and problem solving

1. ## Trignometry - modelling and problem solving

From a point A due north of a tower, the angle of elevation to the top of the tower is 45 degrees. From point B, 100M on a bearing of 120 DEGREES FROM A the angle of elevation is 26 degrees. Find the height of the tower.

Please can you help with this problem

2. ## Re: Trignometry - modelling and problem solving

From a point A due north of a tower, the angle of elevation to the top of the tower is 45 degrees. From point B, 100M on a bearing of 120 DEGREES FROM A the angle of elevation is 26 degrees. Find the height of the tower.

Please can you help with this problem

From the image above Given: $\displaystyle \angle EAD = 45^\circ$, $\displaystyle \angle BAD = 120^\circ$ and $\displaystyle AB = 100m$

We have to find the height of the tower $\displaystyle ED$

Draw from point $\displaystyle G$ to $\displaystyle CD$, $\displaystyle GH$ which is perpendicular to $\displaystyle CD$ and to $\displaystyle BE$, $\displaystyle GO$ perpendicular to $\displaystyle BE$

In $\displaystyle \triangle ABC$:
\displaystyle \begin{align*}\sin {\angle BAC} =& \frac{BC}{AB}\\ =& \frac{BC}{100}....\text{[because }AB = 100, \angle BAD = 120^\circ \text{ and }\angle BAC = 180^\circ - 120^\circ = 60^\circ\text{]}\end{align*}
$\displaystyle \therefore GH = BC = \frac{100\sqrt{3}}{2}$

In $\displaystyle \triangle GAH$:
$\displaystyle \sin {\angle GAH} = \frac{GH}{AG}$
$\displaystyle \therefore AG = \frac{100\sqrt{3}}{\sqrt{2}}....\text{[} \angle GAH = 45^\circ \text{ and } GH = \frac{100\sqrt{3}}{2}}\text{]}$

In $\displaystyle \triangle ABG$:
\displaystyle \begin{align*}BG =& \sqrt{AB^2 + AG^2 - 2*AB*AG*\cos {\angle BAG}}...\text{[ by cosine law]}\\=& 136.60...[AB = 100, AG = \frac{100\sqrt{3}}{\sqrt{2}}, \angle BAG = 75^\circ]\end{align*}

In $\displaystyle \triangle BOG$:
$\displaystyle \sin{\angle OBG} = \frac{OG}{BG}$
$\displaystyle \therefore OG = 59.88^\circ...[\angle OBG = 26^\circ \text{ and } BG = 136.60$

In $\displaystyle \triangle BEF$:
\displaystyle \begin{align*}\angle FEB =& 180^\circ - \angle FBE - \angle BFE \\ =& 180^\circ - 26^\circ - 90^\circ\\=& 64^\circ\end{align*}
Now $\displaystyle \angle AGH = \angle GEF$ because $\displaystyle GF$ is parallel to $\displaystyle AD$ and straight line $\displaystyle AE$ intersects them.

In $\displaystyle \triangle AGH$:
$\displaystyle \angle AGH = 180^\circ - \angle GAH - \angle GHA = 45^\circ...[\angle GAH = 45^\circ, \angle GHA = 90^\circ$

Now:
$\displaystyle \angle FEB = \angle BEA + \angle AED$

Because $\displaystyle \angle AED = \angle GEF = \angle AGH = 45^\circ\text{ and } \angle FEB = 64^\circ$:

$\displaystyle \therefore \angle BEA = 19^\circ$

In $\displaystyle \triangle OEG$:
$\displaystyle \sin{\angle GEO} = \frac{OG}{GE}$

Because $\displaystyle \angle BEA = 19^\circ \text{ and } OG = 59.88$
$\displaystyle \therefore GE = 183.92$

In $\displaystyle \triangle GEF$:
$\displaystyle \cos{\angle GEF} = \frac{EF}{GE}$

But $\displaystyle \angle GEF = 45^\circ \text{ and } GE = 183.92$
$\displaystyle \therefore EF = 130.05$

Now:
\displaystyle \begin{align*}ED = EF + FD = EF + BC =& 130.05 + \frac{100\sqrt{3}}{2}\\ =& 216.65\end{align*}

So the height of the tower is $\displaystyle 216.65 \text{ meter}$

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